Answer:
x=8.75
Explanation:
The price x that maximizes profit is the maximum value of the function, and the maximum value of the function is located at a point where the first derivative of the function is equal to zero. The first derivative is:
![P(x) = - 2x^2+35x-99\\P'(x)=-2(2)x^((2-1))+35(1)-0\\P'(x)=-4x+35](https://img.qammunity.org/2020/formulas/mathematics/college/7huw5t7xs10eehyi5ptd0eu5mwwjdbxpoi.png)
Using P'(x)=0:
![0=-4x+35\\4x=35\\x=35/4\\x=8.75](https://img.qammunity.org/2020/formulas/mathematics/college/30tem6vzc9yr2xnumqz1mlq34wtha6qazi.png)
The minimum value of the function is also at a point where the first derivative of the function is equal to zero. To differentiate if x=8. is a minimum or a maximum obtain the second derivative and evaluate it at x=8.75 if the value P''(x)>0 x is minimum and if P''(x)<0 x is a maximum.
![P'(x)=-4x+35\\P''(x)=-4(1)\\P''(x)=-4](https://img.qammunity.org/2020/formulas/mathematics/college/xie5vuyxg46kjkwqwxfp0kmyfqbndeogvs.png)
Evaluating at x=8.75:
![P''(8.75)=-4](https://img.qammunity.org/2020/formulas/mathematics/college/co3q5w2xwh01kmrrm4ps3ag7ztz10d9atr.png)
Therefore, x=8.75 is the maximum value of the function and it is the price that maximizes profit.